It is well known that the image of $G$ the inversion wrt $\omega$ is the foot from $G$ to $EF$, call it $P$. Let the center of $\omega_1=(A_1B_1C_1D_1)$ be $O_1$, then $GO_1O$ are collinear by homothety. Therefore the image of $\omega_1$ is a line $\ell$ perpendicular to $OO_1$, i.e. perpendicular to $OG$, parallel to $EF$. The image of $O_1$ is the reflection of $O$ wrt $\ell$, call it $Q$, then the midpoint of $QO$ lies on $\ell$, call it $S$. Let the radius of $\omega$ and $\omega_1$ be $R$ and $r$. Then $OP/OQ=OO_1/OG=(R-r)/R$, $GS/GP=(OS-OG)/(OP-OG)=(OQ-2OG)/(2OP-2OG)$. Also note that by homothety, $R/r=(OG+R)/(OG)$, and $OG\cdot OP=R^2$, summing up we would get $GS/GP=0.5(R+r)/R$, as desired.

This solution is isomorphic to the above one. Nice problem (and the first geometry problem that I see in ages that requires the concept of inversion in its statement)! Let $\dfrac{GA_1}{GA}=\ldots=\dfrac{GO_1}{GO}=\dfrac{GL}{GF}=\dfrac{GK}{GE}=\lambda<1$. Let $X,Y$ be the midpoints of $KE,LF$ respectively, and let $OG$ intersect $EF$ at $G'$. Let $O_1$ be the center of circle $(A_1B_1C_1D_1)$, and let $OO_1$ intersect that circle again at $S$, and $XY$ at $T$. We need to prove that $OS \cdot OT=R^2$, with $R$ the circumradius of $\omega$. Note that by the homothety centered at $G$ sending quadrilateral $A_1B_1C_1D_1$ to $ABCD$, we infer that $KL \parallel FE$, and so $XY$ is the midline of the trapezium $EKLF$. Now, note that $GT=GG' \cdot \dfrac{GY}{GF}=GG' \cdot (1-\dfrac{FL}{2FG})=GG' \cdot \dfrac{1+\lambda}{2}$ and so $OS \cdot OT=2OO_1 \cdot (OG+GT)=2OO_1 \cdot (OG+\dfrac{GG'(1+\lambda)}{2})=2(1-\lambda)OG (OG+\dfrac{GG'(1+\lambda)}{2})$ Now, note that points $G,G'$ correspond to each other at the inversion wrt $\omega$, and so $OG \cdot GG'=OG \cdot OG'-OG^2=R^2-OG^2,$ and so $OS \cdot OT=2(1-\lambda)OG (OG+\dfrac{GG'(1+\lambda)}{2})=2(1-\lambda)OG^2+(1-\lambda^2)(R^2-OG^2)=(1-\lambda^2)R^2+(\lambda-1)^2OG^2,$ and so we are left to prove that $(1-\lambda^2)R^2+(\lambda-1)^2OG^2=R^2,$ that is $(1-\lambda)OG=\lambda R$. However, this is easy. Let ray $GO$ intersect $\omega$ at point $N$. Then, point $O$ maps to $N$ under the aforementioned homothety, that is $\dfrac{OG}{R}=\dfrac{GB_1}{B_1B}=\dfrac{\lambda}{1-\lambda},$ as desired.

Here is my solution that doesn't require ratio bash. Pretty much all solutions I have read so far (internally, and in AoPS here) involves some form of bashing @_@ Let ray $OG$ intersect line $EF$ and $\omega$ at $X$ and $Y$ respectively. Let $Z$ be the diametrically opposite point of $Y$ in $\omega$. $\textit{Claim 1:}$ $K, Y, L$ are colinear. In particular $KL$ is tangent to $\omega$ at $Y$. $\textit{Proof:}$ Observe that there is a homothety centered at $G$ that sends the complete quadrilateral $ABCDEF$ to $A_1B_1C_1D_1KL$, with scale factor $0<r<1$. Also note that $O$ is being sent to $Z$ in this homothety, hence in order to prove that $K, Y, L$ are colinear, it suffice to prove that $Y$ is being sent to $X$ in this homothety, that is $$\frac{GO}{OZ}=\frac{GY}{YX}$$But this is because $G$ and $X$ are inverses with respect to $\omega$, and since $O$ is the midpoint of $YZ$, we have $$OG\cdot OX=OY^2 \Rightarrow (X,Y;G,Z)=-1 \Rightarrow \frac{GO}{OZ}=\frac{GY}{YX}$$So $K, Y, L$ are colinear, and since $KL\perp OG$ by Brokard's Theorem, then $KL$ must be tangent to $\omega$ at $Y$. $\square$ By Claim 1, the line passing through midpoints of $KE$ and $LF$ is just the perpendicular bisector of $XY$. Let $A'$ be the center of circle $(AXY)$. We will prove that $A'$ and $A_1$ are inverses with respect to $\omega$ by the following two claims. $\textit{Claim 2:}$ $A', A_1, O$ are colinear. $\textit{Proof:}$ It suffice to prove that $A_1$ is on the perpendicular bisector of $AY$. But observe that $$\frac{GO}{OA}=\frac{GO}{OZ}=\frac{GA_1}{A_1A}$$which implies that $OA_1$ is the angle bisector of $\angle GOA$, so since $O, A_1$ are on the perpendicular bisector of $AY$ as well, the points $A', A_1, O$ are colinear. $\square$ $\textit{Claim 3:}$ The points $O, X, A', A$ are concyclic. $\textit{Proof:}$ Observe that $X$ and $Y$ must be distinct points, otherwise $GY=GX$ implies $A_1B_1C_1D_1$ is the same as $ABCD$. But $O$ lies strictly inside $ABCD$ and hence cannot be concyclic to $ABCD$, violating the second condition. Hence $OX\neq OA$, and since $OA'$ is the angle bisector of $\angle AOX$, then the points $O, X, A', A$ are concyclic. $\square$ Claim $2$ and $3$ implies that $A'$ and $A_1$ are inverses with respect to $\omega$, where $A'$ lies on perpendicular of $XY$. This solves the problem, because similarly the inverses of $B_1, C_1, D_1$ with respect to $\omega$ will be on the perpendicular bisector of $XY$, as desired. $\blacksquare$

Here's an amusing Best Inversion Lemma (BIL) solution. First, let $M$ and $N$ be the Miquel points of $A_1B_1C_1D_1$ and $ABCD$ (the former with circle $\omega_1$ centered at $O_1$). Note that $G\stackrel{\omega_1}{\longleftrightarrow} M\text{ }(\spadesuit)$ hence \[\frac{O_1G}{O_1O}=\frac{O_1O}{O_1M}\Longleftrightarrow \frac{OG}{O_1O}=\frac{OM}{O_1M}\text{ and by homothety at }G\text{ we get }M\in\omega.\]Then, the inversion around $\omega$ fixes $M$, sends $\omega_1$ to line $\ell$ and $G$ to $N$. Because of $(\spadesuit)$ by BIL, $M\stackrel{\ell}{\longleftrightarrow}N$ so $\ell$ is the midline of $EF$ and $KL$.

Another length chasing solution (from the official packet): Let $OG$ intersect $EF$ at $J$, and $OG$ intersect $KL$ at $M$. By Brokard's theorem, $O$ is the orthocenter of triangle $EFG$, and that the radius $r$ of $\omega$ satisfies $r^2 = OG\cdot OJ$. W.l.o.g. we set $OG=1$ (and therefore $OJ=r^2$). Notice that $r > 1$. Now let $\gamma$ be the circle through $A_1, B_1, C_1, D_1, O$. By the ratio condition, $\omega$ and $\gamma$ are homethetic with center $G$, with positive ratio. If ray $GO$ intersects $\omega$ at $P$ then $P$ is mapped (by the homothety map) to $O$, from which we know that this homothety ratio is $\frac{GO}{GP}=\frac{1}{1 + r}$. In a similar way we can deduce that $E$ is mapped to $K$, and $F$ is mapped to $L$, i.e. $\frac{GK}{GE} = \frac{GL}{GF} = \frac{1}{1 + r}$. This implies that $KL\parallel EF$, and therefore $\frac{GM}{GJ} = \frac{1}{1 + r}$. This means $OM = OG + GM = 1 + \frac{1}{r+1}\cdot (r^2-1) = r$ (that is, $\omega$ is tangent to $KL$). Meanwhile, let $\gamma$ intersect $OG$ again at $N\neq O$. Then $\frac{GN}{GM} = \frac{1}{1 + r}$, so $ON = OG + GN = 1 + \frac{1}{1 + r}(r - 1) = \frac{2r}{1 + r}$. Let $N'$ be the image of $N$ in inversion about $\omega$, then $ON' = \frac{r(1+r)}{2}$ and therefore $MN' = ON' - OM = \frac{r(1+r)}{2} - r = \frac{r(r - 1)}{2} = \frac{OJ - OM}{2} = \frac{MJ}{2}$. This means $N'$ is the midpoint of $MJ$. Finally, given that the circles $\omega$ and $\gamma$ are homothetic in $G$ and the centre of $\omega$ $O$ passes through $OG$, the centre of $\gamma$ passes through $OG$. This means the image of $\gamma$ under inversion about $\omega$ is a line (namely $\ell$) perpendicular to $OG$ (and hence parallel to $KL$ and $EF$). Since $\ell$ passes through $N$' the midpoint of $MJ$, $\ell$ is the midline of lines $KL$ and $EF$, and the conclusion follows. $\blacksquare$